Properties of Absolute Value

Introduction to properties of absolute value:

Normally absolute value is nothing but if any value regards to its sign in math. For example let us consider the numbers 8 and -8. Absolute value of these numbers is 8. |8| = +-8. Here we are going to learn about the properties of the absolute value. If we know the properties of the absolute value it is easy to do the operations on the absolute vales.

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Properties of Absolute Value:

Non negativity property:

The absolute value of any numbers is greater than 0. There is no negative numbers in this.

For example take any value -9. So the absolute value is 9. Because|9| = +-9. So for -9 and +9 the absolute value is +9.

|x| >= 0

Positive definiteness:

The absolute value of 0 is always 0. |x| = 0 then x = 0 (always)

Normally for an absolute value we have two values. Here the absolute value is 0. So value of this is 0. There is no sign for the value 0.

Multiplicative property:

Multiplication of any two absolute values same as the individual Absolute Value Equations. This mean

|x `xx` y | = |x| `xx` |y|

Example:

|-2 X 3| = |-6| = + 6

|-2| `xx` |3| = +2 `xx` +3 = + 6

So both the values are equal.

Subtraction addition property:

Addition of any two absolute values is always less than its individual addition.

|x + y| `lt=` |x| + |y|

Example:

|5 + -3| = |2| = 2

|5| + |-3| = 5 + 3 = 8

2 `lt` 8

Other Properties of Absolute Value:

Symmetry property:

Symmetry property is nothing but |-x| = |x|

Absolute value of –x and absolute value of x is always equal.

Identity of indiscernible property:

If the subtraction of any two absolute values is 0 then these two absolute values are equal.

|x - y| = 0 then x = y

Preservation of division:

The division of any two absolute values same as individual absolute value division.

|x / y | = |x| / |y| (where y `!=` 0)

These are the some basic properties of the absolute values.

Least Common Multiple and Greatest Common Divisor

Introduction to least common multiple and greatest common divisor
Least common multiple

In mathematics, the least common multiple (lcm), of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder.

Greatest common divisor

In mathematics, the greatest common divisor (gcd), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. (Source: From Wikipedia).

Example Problems to Find the least Common Multiple and Greatest Common Divisor

Example problems to find the least common multiple of two numbers

Example 1

Find the least common multiple of 3 and 7.

Solution

To find the least common multiples of 3 and 7, first we have to find the multiples of 3 and 7.

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70

Here, 21 is the lowest common number in the multiples of 3 and 7.

So, 21 is the least common multiple of 3 and 7.

Example 2

Find the least common multiple of  5, 11

Solution

The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70.

The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132.

Here, 55 is the lowest common number in the multiples of 5 and 11.

So, 55 is the least common multiple of 5 and 11.

Example problems to find the greatest common divisor of two numbers.

Example 1

Find the greatest common divisor of  54, and 40

Solution

The greatest common divisor of 54 and 40 can be found by the prime factors of the given numbers.

Prime factorization of 54 = 2 * 3 * 3 * 3

Prime factorization of 40 = 2 * 2 * 2 * 5

The common factors in the prime factorization is 2

So, 2 is the greatest common divisor of 54 and 40.

Example 2

Find the greatest common divisor of 21 and 35

Solution

Prime factorization of 21 = 3 * 7

Prime factorization of 35 = 5 * 7

Here, 7 is the common term.

So, 7 is the greatest common divisor of 21 and 35.

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Practice Problems to Find the least Common Multiple and Greatest Common Divisor

Problem 1

Find the greatest common divisor of 3, 5, 7

Answer: 1

Problem 2

Find the greatest common divisor of 54, 27

Answer: 27

Problem 3

Find the least common multiple of 21 and 3

Answer: 21

Problem 4

Find the least common multiple of 51 and 16.

Answer: 816

Finding Percentages

Introduction :
Percentages are numerators of fractions with denominator 100 and have been used in comparing results. Per cent is derived from Latin word ‘per centum’ meaning ‘per hundred’. Per cent is represented by the symbol % and means hundredths too. That is 1% means 1 out of hundred or one hundredth. It can be written as: 1% =1/100 =0.01. We saw how percentages were helpful in comparison. We have also learn to convert fractional numbers and decimals to percentages. Now, we shall learn how percentages can be used in real life.

Finding Percentages for Fractions

Fractional numbers can have different denominator. To compare fractional numbers, we need a common denominator and we have seen that it is more convenient to compare if our denominator is 100. That is, we are converting the fractions to Percentages. Let us try converting different fractional numbers to Percentages.

Example: Write 1/3 as per cent.

Solution:

We have, 1/3 = 1/3 x 100/100 = 100 %/3=33 and 1/3 %

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Finding Percentages for Decimals

We have seen how fractions can be converted to percents. Let us now find how decimals can be converted to percents.

Example1: Convert the given decimals to per cents:

(a)    0.75 (b) 0.09 (c) 0.2

Solution:

(a)    0.75 = 0.75 × 100 % =(75/100)× 100 % = 75%

(b)   0.09 =9/100= 9 %

(c)    0.2 =(2/10) × 100% = 20 %

Example2: The monthly salary of Meena is Rs. 4000. She spends 80% of her salary every month. How much does she save every month?

Solution:

Meena's monthly salary = Rs. 4000

Expenditure = 80% of 4000 = 80/100 × 4000 = Rs. 3200

Therefore, Savings = 4000 – 3200 = Rs. 800

Prime Numbers to 1000

Introduction to prime numbers to 1000:

A natural number other than 1, divisible only by itself and 1. Or any natural number which has precisely two divisors. The following numbers are prime numbers are prime 2, 3, 5, 7, 11, 13 … 37… every natural numbers greater than 1 may be resolved uniquely into a product of prime numbers. For example 180 = 2 * 2 * 3 * 3 * 5. In the case of prime number p, the product has to be interpreted as p itself.

Finding Prime Number to 1000:

Here we will see how to find the prime numbers. Which two types of prime, Both Fermat and Mersenne have presented us a few techniques to identify the prime numbers. Neither always gets up a prime number, but in some time of Mersenne, his formula leads to finding exact numbers.

Fermat’s formula: 2n + 1

Messene’s formula: 2n – 1

We do not know whether there are a countless number of Mersenne primes.

Or whether we can get the infinite number of primes from Fermat’s Formula.

Some facts:

The even prime number is 2 only. Remaining even numbers can be divided by 2.
If the total of a number's digits are multiple by 3, which number can be divided by 3.
No prime number < 5 ends in a 5. Several numbers < 5 which ends in a 5 can be divided by 5.
0 and 1 are not measured as a prime numbers.
Apart from for 0 and 1, a number is whether a prime number or a composite number. A composite number is distinct as any number, < 1, which is not a prime.
To demonstrate either a number is a prime number, initially try to dividing it by 2 and look if we get a whole number. If we do, it cannot be a prime number. If we do not get a whole number, then try to dividing it by prime numbers: 3, 5, 7, and 11 (9 are divisible by 3) etc, Let us see some prime numbers from 1 to 1000 in the below table.

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Example for Prime Numbers to 1000:

The following is the list of prime numbers below 1000.

2  3  5  7  11  13  17  19  23  29  31  37  41  43

47  53  59  61  67  71  73  79  83  89  97  101  103  107

109   113  127   131   137   139   149   151   157   163   167   173   179   181

191   193   197   199   211   223   227   229   233   239   241   251   257   263

269   271   277   281   283   293   307   311   313   317   331   337   347   349

353   359   367   373   379   383   389   397   401   409   419   421   431   433

439    443  449   457   461   463   467   479   487   491   499   503   509   521

523    541    547    557    563    569   571   577   587   593   599   601   607   613

617    619     631    641    643    647   653   659    661    673    677    683    691   701

709   719    727    733    739    743    751    757    761    769     773     787    797    809

811    821    823    827     829     839    853    857    859    863    877    881    883    887

907    911    919    929    937   941   947   953   967   971   977   983    991   997

Methods of Factoring Trinomials

Introduction :

Trinomials:

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.

Trinomial expressions:

1. 3x + 5y + 8z with x, y, z variables

2. 3t + 9s2 + 3y3 with t, s, y variables

3. 3ts + 9t + 5s with t, s variables

4. Axaybzc + Bt + Cs with x, y, z, t, s variables, a, b, c nonnegative integers and A, B, C any constants.

5. Pxa + Qxb + Rxc where x is variable and constants a,b,c are nonnegative integers and P, Q, R any constants.

6. x2 + 8x + 15 where x is variable ( source : wikipedia)

Methods of Factoring Trinomials:

By using the following two methods we can factoring the trinomials.

Factoring trinomials Method 1:

The co-efficient of the first term (that is x2) is one. That is a=1.

x2+bx+c=(x-r1)(x-r2), Here r1 and  r2 are the roots of the trinomials.

We can say, (x - r1), (x - r2) are the factors of the trinomial.

Example Problems on methods of factoring trinomials :

By go through the following problem you can learn the method 1 of factoring the trinomials.

Example 1:

Factor the trinomial x2- 10 x +16

Solution:

Given, x2- 10 x +16

We need to factoring the given trinomial,

16   (product)

/    \   

- 8      - 2

\    /

-10     (sum)

So we can write the given equation into,

x2- 10 x +16 = x2 - 2x - 8x +16

= x ( x-2 ) - 8( x-2 )

= ( x - 8 ) ( x - 2 )

Answer: (x-8) and (x-2) are the factors of the given trinomial.

Verification:

(x-8) (x-2)  = x (x - 8) -2 (x - 8)

= x2-8x -2x+16

(x-8) (x-2)  = x2 -10x + 16

Example 2

Find the factors of   x2+ 3x -18.

Solution:

Given, x2+ 3x -18

We need to factoring the given trinomial,

-18 ( Product)

/     \

6       - 3

\     /

3 ( sum)

So we can rewrite the given equation into,

x2+3x - 18 = x2-3 x+6x -18

= ( x2-3x ) + ( 6x-18 )

= x ( x-3 ) + 6 ( x -3 )

= (x - 3) (x + 6)

Answer: factors (x-3) and (x+6)

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Methods of Factoring Trinomials:

Factoring trinomials Method 2:

The co-efficient of the first element( x2) trinomial is greater than one. That is a > 1.

The following example will help you to understand this method.

Example:

Factoring the trinomial 6x2- 3x - 3.

Solution:

Given, 6x2- 3x - 3 

To factor the trinomial, multiply the coefficient of first term with the constant i.e. 6 * (-3)=-18

- 18    (product)

/    \

- 6     3

\     /

- 3    (sum)

6x2- 3x - 3  =  6x2- 6x + 3x -3

=  6x (x - 1) + 3(x-1)

= (6x+3) (x-1)

= 3(2x+1)(x-1)

Answer: Factors of the given trinomial 3, (2x+1), (x-1)

Easy way to solve mean

Mean is nothing but average value of set of numbers. Mean almost always refers to arithmetic mean.

Mean = Total of all the elements in a data set / Total number of elements in a data set

The following steps are given below:

Step 1: Find how many elements are there in a set.

Step 2: Find the sum of all the elements in a set.

Step 3: Divide the sum by total number of elements in a set.

Step 4: Final answer is the required mean.

Easy Way to Solve Mean - Examples

Example 1: Solve and find the arithmetic mean value for following data {32, 45, 59, 62, 19, 72, 21, 42, 87}

Solution:

Step 1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

32 + 45 + 59 + 62 + 19 + 72 + 21 + 42 + 87 = 439

Step 3: Divide the sum by total number of elements in a set

= Sum of all the element / Total number of elements.

= 439 / 9

= 48.77

Therefore arithmetic mean of given date is 48.77

Example 2: Solve and find the arithmetic mean for following data {21, 42, 63, 74, 56, 46, 57, 68, 79}.

Solution:

Step1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

21 + 42 + 63 + 74 + 56 + 46 + 57 + 68 + 79 = 506

Step 3: Divide the sum by total number of elements in a set

= Sum of all the element / Total number of elements.

= 506 / 9

= 56.22

Therefore arithmetic mean of given date is 56.22

Example 3: Calculate the arithmetic mean for following data {14, 29, 37, 43, 26, 52, 62, 71, 24}.

Solution:

Step1: Find how many elements are there in a set.

There are 9 elements, so n = 9.

Step 2: Find the sum of all the elements in a set.

14 + 29 + 37 + 43 + 26 + 52 + 62 + 71 + 24 = 358

Step 3: Divide the sum by total number of elements in a set.

= Sum of all the element / Total number of elements.

= 358 / 9

= 39.77

Therefore arithmetic mean of given date is 39.77

Example 4: Calculate the arithmetic mean for following data {31, 43, 68, 87, 85, 69, 157, 277}.

Step1: Find how many elements are there in a set.

There are 8 elements, so n = 8.

Step 2: Find the sum of all the elements in a set.

31 + 43 + 68 + 87 + 85 + 69 + 157 + 277 = 817

Step 3: Divide the sum by total number of elements in a data set.

= Sum of all the element / Total number of elements.

= 817 / 8

= 102.125

Therefore arithmetic mean of given date is 102.125

Easy Way to Solve Mean - Practice

Solve these practice problems using easy way. These practice problems are very easy to solve

Problem 1: Calculate the arithmetic mean for following data {66, 55, 97, 88, 35, 36, 74, 82}.

Answer: 66.62

Problem 2: Calculate the arithmetic mean for following data {232, 346, 445, 556, 662, 771, 774, 321}.

Answer: 513.375

An Introduction to Regression Analysis

Let’s have an example of regression .A new plant to manufacture widgets was set up, the plant’s personnel manager advertises the employment opportunity in the plant and the very next morning it has 2000 people waiting for applying to the 100 available jobs. It will be important to select the best of 100 people who will cut above the rest employees because the training of these people will involve both time and money and firing the employee is going to be difficult as well as bad for community relations. In order to overcome this situation and to help make correct decision an analysis is adopted by the Personnel manager, this analysis is the Regression analysis.  A Regression analysis is used to predict the purposes and to understand relationship between two variables x and y. It is used when two or more variables are systematically connected by a linear relationship.  A Multivariate Regression Analysis is a technique that estimates a single regression model with more than one outcome variable.

In a Regression equation, y is the dependent variable and x is a independent variable, the Regression Formula can be given as y’= a +bx; where y’ is the estimated y which is on the y-axis from the point on the regression line for the predictor x value, x is an arbitrary chosen value of the predictor variable for which the corresponding value of the criterion variable is desired, a is the intercept point of the regression line and the y-axis and b is the slope of the regression line. a and b are calculated as follows,
a = [sigma(y)][sigma(x2)] – [sigma(x)][sigma(x)(y)]divided by {n [sigma(x2)]- [sigma(x)]2} and

b= n[sigma(xy)] – [sigma(x)][sigma(y)] divided by {n [sigma(x2)]- [sigma(x)]2}

Regression Analysis Example or Regression Example: Regression analysis is an analysis to assess the values of the given parameters for a particular function that cause the function to the best utilize a set of observed datas that are given. For example, if we take into account the value of an automobile, it decreases constantly with a certain amount each year after the purchase, and each for mile it has driven, the related linear function would predicts its value as a function of the two independent variables like ‘age’ and ‘miles’:  value = price + dep age x age + dep miles x miles; where value is the dependent variable which is the value of the car, age is the age of the car and the miles is the number of miles that the car has covered. The regression analysis done determines the best of values of the three parameters, price, the estimated value when the age is zero, dep-age, which is the depreciation that takes place each year, dep-miles is the depreciation for each mile driven.  The values of dep-age and dep-miles will be -ve as the car would be losing value as the car age and the miles covered increase.